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Viefleur [7K]
3 years ago
12

What steps would you take to design an improved toothpaste container?

Engineering
1 answer:
algol [13]3 years ago
7 0

Answer:

A. Identify the need, recognize limitations of current toothpaste containers, and then brainstorm ideas on how to improve the existing

Explanation:

To design an improved toothpaste container, we must identify the needs of the customer, one of the major need is to make the container attractive to the sight. This is the first thing that will prompt a customer to wanting to buy the product (The reflectance/appearance).

Then recognize the limitation of the current design, what needed change. This will help in determining what is needed to be included and what should be removed based on identified customers need.

The last step is to brainstorm ideas on how to improve the existing designs. Get ideas from other colleagues because there is a saying that two heads are better than one. This will help in coming to a reasonable conclusion on the new design after taking careful consideration of people's opinion.

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A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
4 years ago
EJERCICIO 6
Ket [755]

Answer:

mnfokfnfi3or

Explanation:

can you translate it into english.....

8 0
3 years ago
To produce cooling the refrigerant much
lawyer [7]
Yes that is correct good job ❤️
7 0
4 years ago
The heat rate is essentially the reciprocal of the thermal efficiency. a)- True b)- False
jeyben [28]

Answer:

a). TRUE

Explanation:

Thermal efficiency of a system is the defined as the ratio of the net work done to the total heat input to the system. It is a dimensionless quantity.

Mathematically, thermal efficiency is

        η =  net work done / heat input

While heat rate is  the reciprocal of efficiency. It is defined as the ratio of heat supplied to the system to the useful work done.

Mathematically, heat rate is

       Heat rate = heat input / net work done

Thus from above we can see that heat rate is the reciprocal of thermal efficiency.

Thus, Heat rate is reciprocal of thermal efficiency.

4 0
3 years ago
Consider two different flows over geometrically similar airfoil shapes, one airfoil being twice the size of the other. The flow
yulyashka [42]

Answer

No;

The two flows are not dynamically similar

 Explanation:  Given

T∞,1 = 800k

V∞,1 = 200m/s

p∞,1 = 1.739kg/m³

T∞,2 = 200k

V∞,2 = 100m/s

p∞,2 = 1.23kg/m³

Size1 = 2 * Size2 (L1 = 2L2)  Assumptions Made

α ∝√T

μ∝√T  Two (2) conditions must be met if the two flows are to be considered similar.

Condition 1: Similar Parameters must be the same for both flows

Condition 2: The bodies and boundaries must be genetically true.  Condition 2 is true

Checking for the first condition...

Well need to calculate Reynold's Number for both flows

And Check if they have the same Reynold's Number  Using the following formula

Re = pVl/μ

Re1 = p1V1l1/μ1  Re2 = p2V2l2/μ2  Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2

Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2

Re1/Re2 = p1V1l1μ2/p2V2l2μ1

Re1/Re2 = p1V1l1√T2 / p1V1l1√T1

Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)

Re1/Re2 = 9837.2/3479

Re1/Re2 = 2,828/1

Re1:Re2 = 2.828:1

Re1 ≠ Re2,

So condition 1 is not satisfied  Since one of tbe conditions is not true, the two flows are not dynamically similar

5 0
3 years ago
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